1889 Iowa Senate election

1889 Iowa Senate election

← 1887 November 5, 1889 1891 →

22 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party Third party
 
Party Republican Democratic Independent
Last election 32 16 2
Seats after 28 20 1
Seat change Decrease4 Increase4 Decrease1

  Fourth party
 
Party Union Labor
Last election 0
Seats after 1
Seat change Increase1

     Democratic hold      Republican hold
     Democratic gain      Union Labor gain

In the 1889 Iowa State Senate elections Iowa voters elected state senators to serve in the twenty-third Iowa General Assembly. Elections were held in 22 of the state senate's 50 districts. State senators serve four-year terms in the Iowa State Senate.

A statewide map of the 50 state Senate districts in the 1889 elections is provided by the Iowa General Assembly here.

The general election took place on November 5, 1889.[1]

Following the previous election, Republicans had control of the Iowa Senate with 32 seats to Democrats' 16 seats and two Independents.

To claim control of the chamber from Republicans, the Democrats needed to net 10 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1889 general election with the balance of power shifting to Republicans holding 28 seats, Democrats having 20 seats, one Independent, and one member of the Union Labor Party (a net gain of 4 seats for Democrats and 1 seat for the Union Labor Party).

  1. ^ "Roll of Members of the Twenty-Third General Assembly. SENATE" (PDF). Iowa Official Register. Retrieved June 26, 2021.

© MMXXIII Rich X Search. We shall prevail. All rights reserved. Rich X Search